Find the mean of the following frequency dist | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To find the mean,we use the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$,where $x_i$ is the class mark.$1$. Calculate class marks $(x_i)$: $5

Vedclass · Accepted Answer

$47.2$

Vedclass · Answer

(C) To find the mean,we use the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$,where $x_i$ is the class mark.$1$. Calculate class marks $(x_i)$: $5, 15, 25, 35, 45, 55, 65, 75$.$2$. Calculate $f_i x_i$: $2 	imes 5 = 10$$4 	imes 15 = 60$$10 	imes 25 = 250$$20 	imes 35 = 700$$18 	imes 45 = 810$$20 	imes 55 = 1100$$16 	imes 65 = 1040$$10 	imes 75 = 750$$3$. Sum of frequencies $(\sum f_i)$ = $2 + 4 + 10 + 20 + 18 + 20 + 16 + 10 = 100$.$4$. Sum of $f_i x_i$ $(\sum f_i x_i)$ = $10 + 60 + 250 + 700 + 810 + 1100 + 1040 + 750 = 4720$.$5$. Mean $\bar{x} = \frac{4720}{100} = 47.2$.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$2$	$4$	$10$	$20$	$18$	$20$	$16$	$10$

Literacy rate (in $\%$)	$45-55$	$55-65$	$65-75$	$75-85$	$85-95$
Number of cities	$3$	$10$	$11$	$8$	$3$

Class	$200-299$	$300-399$	$400-499$	$500-599$	$600-699$	$700-799$	$800-899$
Frequency	$3$	$61$	$118$	$139$	$126$	$151$	$2$

Find the mean of the following frequency distribution:

Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$

Frequency $2$ $4$ $10$ $20$ $18$ $20$ $16$ $10$

Similar Questions

The following table gives the literacy rate (in percentage) of $35$ cities. Find the mean literacy rate.
Literacy rate (in $\%$) $45-55$ $55-65$ $65-75$ $75-85$ $85-95$
Number of cities $3$ $10$ $11$ $8$ $3$

Find the mean of the following data:

Class $200-299$ $300-399$ $400-499$ $500-599$ $600-699$ $700-799$ $800-899$

Frequency $3$ $61$ $118$ $139$ $126$ $151$ $2$

If $x_{i}$ are the midpoints of the class intervals of grouped data,$f_{i}$ are the corresponding frequencies,and $\bar{x}$ is the mean,then $\sum (f_{i} x_{i} - f_{i} \bar{x})$ is equal to

The mean of $15$ observations is $16$. If $2$ is added to each observation and then each result is divided by $3$,the new mean is:

In the formula $\bar{x} = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}}$ for finding the mean of grouped data,$d_{i}$ are deviations from $a$ of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find the mean of the following frequency distribution: Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ Frequency $2$ $4$ $10$ $20$ $18$ $20$ $16$ $10$